Physical Optics and Diffraction
Simone Ferraro
Princeton University
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Outline
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Physical optics and Kirchhoff Integral
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Diffraction by an aperture
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Fraunhofer diffraction
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Fresnel diffraction
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Image formation
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Dealing with imperfections
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Kirchhoff integral
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Monochromatic scalar wave
with spatial part satisfying Helmholtz eqn:
for free propagation, eg: one component of
E field in absence of polarization coupling
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Assume medium is homogeneous and nondispersive so that k is constant
Helmholtz eqn is a linear, elliptic PDE, so
solution inside a volume is completely
determined by value of
and its normal
derivative on the boundary.
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Kirchhoff integral (cont..)
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Nice trick available!
Greens' theorem: for any two (reasonable)
scalar functions
boundary
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Above intergrals = 0 if both functions
satisfy Helmholtz equation
By inspection
is a solution
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Kirchhoff integral (cont..)
=0
boundary
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With
, have
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So integral over S0 becomes
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Therefore
S
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Diffraction by an aperture
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Suppose we have an aperture, which is
big compared to the wavelength, but small
compared to the distance to .Illumination
comes from a distant wave source.
Characterize aperture by a complex
function , such that the wave just after
passing through it
On the aperture have
,
, with
, so write
pointing towards
parallel to k
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Diffraction by an aperture
Then
Can solve in two different regimes:
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Far field (Fraunhofer)
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Near field (Fresnel)
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Fraunhofer regime (far field)
Condition for Fraunhofer is that we can
neglect the quadratic phase variation with
position in the aperture plane
i.e.
for all x, or
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Fraunhofer regime (far field)
Neglecting the quadratic term:
The amplitude of the Fraunhofer diffraction
pattern is given by the 2D Fourier transform of
the aperture function
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Fraunhofer examples
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Diffraction by a single, finite slit
Then
Intensity
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Fraunhofer examples
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Diffraction grating
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Fraunhofer examples
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Diffraction grating
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p-th order beam deflected by
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Consider beams for waves at
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Located at
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Separation
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and
Can distinguish them if maximum of one
corresponds to first minimum of the other,
i.e.
is at least
(Rayleigh criterion)
Corresponds to
power)
(chromatic resolving
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Fraunhofer examples
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Diffraction by a circular aperture
Most of the light from a distant
source falls within the Airy disc
Can use to calculate the
diffraction limit of a lens/telescope
Two equally bright sources can be resolved only
if the radius of the Airy disk is less than their
separation, i.e if their angular separation is more
than
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Fraunhofer examples
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Diffraction by a circular aperture
Two equally bright sources can be resolved only
if the radius of the Airy disk is less than their
separation, i.e if their angular separation is more
than
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Fresnel regime (near field)
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Now can't neglect quadratic phase variation
Problem a bit harder. Consider special case in which
the source, the origin of coordinates and the
observation point are aligned
O
Might seem
uninteresting, but
can make a lot of
progress by moving
the origin!
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Fresnel regime (near field)
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Fresnel regime (near field)
write
then
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Separable aperture
If the aperture function is separable (e.g.
rectangular aperture), it is convenient to rewrite in
terms of Fresnel integrals.
Change variables to
Define Fresnel integral
Plotting C vs S, obtain the
Cornu spiral
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Separable aperture
If the aperture function is separable (e.g.
rectangular aperture), it is convenient to rewrite in
terms of Fresnel integrals.
Change variables to
Define Fresnel integral
Plotting C vs S, obtain the
Cornu spiral
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Diffraction from
straight
edge
x
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Diffraction from straight edge
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Use the Fresnel pattern from lunar
occultation of the powerful 3C 273 radio
source to determine its position accurately
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Diffraction by circular aperture
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Consider circular aperture w/ radius D
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Retain the obliquity factor K
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Using polar coordinates:
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Use the substitution
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Note: only valid on axis!
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Diffraction by circular aperture
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Evaluate graphically
Going around in a
circle
Spiralling in
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Diffraction by circular aperture
and Fresnel Zones
For finite apertures, the diffraction integral varies
considerably
Define Fresnel zones as
concentric rings in the
aperture plane over
which the phase varies at
the observation point
varies by
1st zone:
Nth zone:
Note: Area of each zone is the same
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Diffraction by circular aperture
and Fresnel Zones
For finite apertures, the diffraction integral varies
considerably
Define Fresnel zones as
concentric rings in the
aperture plane over
which the phase varies at
the observation point
varies by
1st zone:
Nth zone:
Note: Area of each zone is the same
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Diffraction by circular aperture
and Fresnel Zones
For finite apertures, the diffraction integral varies
considerably
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Zone plate
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A Fresnel zone plate is an aperture that blocks
alternates half-period zones.
Eg: block the odd ones
Net amplitude at P:
Intensity
Incident wave brought to focus at P!
Acts like a lens of focal length
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Zone plate (cont..)
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→ highly chromatic lens
Focuses different wavelengths in different
foci
Works for any frequency (including X-rays)
Maximum resolution depends on smallest
zone width
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Image formation (thin lens)
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Field at distance zj along the optical axis:
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Vector x is perpendicular to the optic axis
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For a linear system
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P21 is the propagator or Point Spread Function (PSF)
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Image formation (thin lens)
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Field at distance zj along the optical axis:
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Vector x is perpendicular to the optic axis
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For a linear system
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P21 is the propagator or Point Spread Function (PSF)
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Free propagation through
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Thin lens: phase shift depends quadratically on the
distance from optic axis |x|
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Image formation (thin lens)
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Propagating to the focal plane:
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Image formation (thin lens)
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Propagating from source to the focal plane:
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Amplitude given by
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Therefore the amplitude in the focal plane is
proportional to the Fourier transform of the field in
the source plane
Focal plane can be used for spatial filtering, i.e.
processing the image by altering its Fourier
transform
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Image formation (thin lens)
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Finally, free propagation from focal plane
to image plane
Apart from a phase factor, the field in the
image plane is just a magnified version of
the original field, with the correct
magnification
Theory due to E. Abbe (1873)
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Dealing with imperfections...
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So far only considered perfect lenses/mirrors and
uniform propagation medium
When any of the above assumptions fail, the image
quality is degraded
Useful to define the Strehl ratio as the ratio
between the peak amplitude of the actual PSF and
the peak amplitude expected in the presence of
diffraction only
Looking through the atmosphere, the Strehl ratio will
be very low, even with good optics (due to
atmospheric turbulence).
In presence of Adaptive Optics systems, can get
very close to 1 (diffraction limited)
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Dealing with imperfections...
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Can treat aberrations and imperfections in
the physical optics language (but it's a
hard problem, see Born & Wolf)
Can relate rms imperfection in the optics
with actual intensity in the focal plane
For example, in order to have
we need
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References
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Born & Wolf, Principles of Optics
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Hecht, Optics
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Blanford & Thorne, Applications of
Classical Physics lecture notes
Part IB Optics lecture notes (Cambridge)
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